A population has parameters u = 239.9 and o = 18.4. You intend to draw a random sample of size n = 76. What is the mean of the distribution of sample means? What is the standard deviation of the distribution of sample means? (Report answer accurate to 2 decimal places.)
A population has parameters u = 239.9 and o = 18.4. You intend to draw a random sample of size n = 76. What is the mean of the distribution of sample means? What is the standard deviation of the distribution of sample means? (Report answer accurate to 2 decimal places.)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
![**Distribution of Sample Means**
A population has parameters \(\mu = 239.9\) and \(\sigma = 18.4\). You intend to draw a random sample of size \(n = 76\).
### What is the mean of the distribution of sample means?
\[
\mu_{\bar{x}} = \text{ }
\]
### What is the standard deviation of the distribution of sample means?
(Report answer accurate to 2 decimal places.)
\[
\sigma_{\bar{x}} = \text{ }
\]
In this exercise, you need to determine both the mean and standard deviation of the distribution of sample means for the given population parameters and sample size.
### Steps to Solve:
1. **Mean of the Distribution of Sample Means**:
The mean of the distribution of sample means \(\mu_{\bar{x}}\) is equal to the mean of the population \(\mu\).
\[
\mu_{\bar{x}} = \mu
\]
2. **Standard Deviation of the Distribution of Sample Means**:
The standard deviation of the distribution of sample means \(\sigma_{\bar{x}}\) is calculated using the population standard deviation \(\sigma\) and the sample size \(n\).
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
Substitute the given values:
\[
\sigma_{\bar{x}} = \frac{18.4}{\sqrt{76}}
\]
By following these steps, you can complete the blanks in the exercise.
There are no graphs or diagrams to explain for this particular content. The central focus is on applying the formulas to find the required statistical measures given the population parameters and sample size.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa218d9b0-33cd-49fb-8966-615774819802%2Fe64d77c8-73df-4c23-abb3-71b4334a1e98%2Fxb9709_processed.png&w=3840&q=75)
Transcribed Image Text:**Distribution of Sample Means**
A population has parameters \(\mu = 239.9\) and \(\sigma = 18.4\). You intend to draw a random sample of size \(n = 76\).
### What is the mean of the distribution of sample means?
\[
\mu_{\bar{x}} = \text{ }
\]
### What is the standard deviation of the distribution of sample means?
(Report answer accurate to 2 decimal places.)
\[
\sigma_{\bar{x}} = \text{ }
\]
In this exercise, you need to determine both the mean and standard deviation of the distribution of sample means for the given population parameters and sample size.
### Steps to Solve:
1. **Mean of the Distribution of Sample Means**:
The mean of the distribution of sample means \(\mu_{\bar{x}}\) is equal to the mean of the population \(\mu\).
\[
\mu_{\bar{x}} = \mu
\]
2. **Standard Deviation of the Distribution of Sample Means**:
The standard deviation of the distribution of sample means \(\sigma_{\bar{x}}\) is calculated using the population standard deviation \(\sigma\) and the sample size \(n\).
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\]
Substitute the given values:
\[
\sigma_{\bar{x}} = \frac{18.4}{\sqrt{76}}
\]
By following these steps, you can complete the blanks in the exercise.
There are no graphs or diagrams to explain for this particular content. The central focus is on applying the formulas to find the required statistical measures given the population parameters and sample size.
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